What are the solutions to the equation x^2+16x+128=0 ?

The solutions to the equation x^2+16x+128=0 are x=-8+8i,x=-8-8i

Find the zeros of x^2+16x+128=0

The zeros of x^2+16x+128=0 are x=-8+8i,x=-8-8i

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x^{2} + 16 x + 128 = 0

x = - 8 + 8 i, x = - 8 - 8 i

Solution steps

x^{2} + 16 x + 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 128}{2 \cdot 1}

Simplify

1 6^{2} - 4 \cdot 1 \cdot 128 : 16 i

x_{1, 2} = \frac{- 16 \pm 16 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 16 + 16 i}{2 \cdot 1}, x_{2} = \frac{- 16 - 16 i}{2 \cdot 1}

x = \frac{- 16 + 16 i}{2 \cdot 1} : - 8 + 8 i

x = \frac{- 16 - 16 i}{2 \cdot 1} : - 8 - 8 i

The solutions to the quadratic equation are:

x = - 8 + 8 i, x = - 8 - 8 i

4 (3 x + 8)^{2} + 17 = 21

y, 2 (x^{2} + y^{2}) = 25 (x^{2} - y^{2})

12 x^{2} - 5 x - 2 = 0

s^{2} + 6 s + 13

2 v^{2} + 4 v = - 10

What are the solutions to the equation x^2+16x+128=0 ?
The solutions to the equation x^2+16x+128=0 are x=-8+8i,x=-8-8i
Find the zeros of x^2+16x+128=0
The zeros of x^2+16x+128=0 are x=-8+8i,x=-8-8i